Reaction favorability and Keq... HELP!!

[pineappletree]

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I was reading through the BR book and I noticed that it said
delta G<0 means Keq>1

delta G>0 means Keq<1


I am a little confused. How can reactions with delta G>0 have a Keq at all? I thought they were unfavorable and hence unable to react~~
I know that you can add temperature to move the delta G down, but then it wouldn't really be delta G>0 anymore now wouldn't it?

 

[muhali3]

Unfavorable but still do react, its just that the reverse reaction occurs more than the forward reaction, thus Keq < 1....that's how i learned it, not 100% sure.

 

[pineappletree]

haha, that erases sooo much I had previously thought of delta G as



So what exactly is the different between positive and negative delta G if both require an energy input and both are able to react?
Are they essentially the same thing except with a varying degree of energy input required??

This would mean that the terms "Favorable" and "Unfavorable" are just arbitrary terms slotted to define a relative amount of Keq at <1 and >1??

 

[lesset01]

Delta G refers to Gibbs free energy which is a measure of whether or not the reaction will spontaneously occur in nature. So if a rxn occurs spontaneously (delta G < 1), it makes sense that its Keq would be > 1 b/c it would favor products.

If a rxn does not occur spontaneously then its Keq would be < 1 and would favor reactants. That doesn't mean that the rxn doesn't occur, just that products are not favored and that the reaction does not spontaneously occur in nature w/o some type of outside influence put on the system (ie increase in temp).

 

[Phantastic]

Woah, woah, woah. Hold your horses peeps!!!

Keq is equilibrium concentrations of products / reactants.

At equilibrium, the forward reaction rate is equal to the reverse reaction rate...therefore delta G = 0 at equilibrium. This is an important fact!

What they were likely referring to in BR, was delta Go (pronounced delta G "not").

delta G = delta Go + RTlnQ

...at equilibrium delta G = 0....

0 = delta Go + RTKeq

delta Go = -RTKeq

So, when:

delta Go < 0; Keq > 1
delta Go > 0; Keq < 1
delta Go = o; Keq = 1

so, whereas delta G is a statement of spontaneity, delta Go is more of a statement of product/reactant concentrations at equilibrium.

Careful with the logic...it's easy to presume that -delta G would favor products over reactants at equilibrium. But that's not the case.

Assume: X + Y -----> Z

and assume that Keq = 0.00000000001

If I start with just a bunch of X and Y in my solution, you can compute delta G to be very negative, yet Keq is not greater than 1. See?

 

[lesset01]

Thanks for clarifying! I just threw that out off the top of my head. I guess I should have checked a reference 1st!

Woah, woah, woah. Hold your horses peeps!!!

Keq is equilibrium concentrations of products / reactants.

At equilibrium, the forward reaction rate is equal to the reverse reaction rate...therefore delta G = 0 at equilibrium. This is an important fact!

What they were likely referring to in BR, was delta Go (pronounced delta G "not").

delta G = delta Go + RTlnQ

...at equilibrium delta G = 0....

0 = delta Go + RTKeq

delta Go = -RTKeq

So, when:

delta Go < 0; Keq > 1
delta Go > 0; Keq < 1
delta Go = o; Keq = 1

so, whereas delta G is a statement of spontaneity, delta Go is more of a statement of product/reactant concentrations at equilibrium.

Careful with the logic...it's easy to presume that -delta G would favor products over reactants at equilibrium. But that's not the case.

Assume: X + Y -----> Z

and assume that Keq = 0.00000000001

If I start with just a bunch of X and Y in my solution, you can compute delta G to be very negative, yet Keq is not greater than 1. See?